Question

A photoelectric surface is illuminated successively by monochromatic light of wavelength ‘(λ)’ and ‘(\(\frac {λ}{2}\))'. If the maximum kinetic energy of the emitted photoelectrons in the first case is one-third that in the second case, the work function of the surface of the materialis (c = speed of light, h = Planck’s constant.)

• A. \(\frac {hc}{3λ}\)
• B. \(\frac {hc}{2λ}\)
• C. \(\frac {2hc}{λ}\)
• D. \(\frac {hc}{λ}\)

Answer 1

The Correct Option is B

Let's denote the work function of the surface of the material by φ. The maximum kinetic energy (KEmax) of the emitted photoelectrons can be determined using the equation: 
KE_max = hν - φ 
In the first case, the incident light has a wavelength of λ. The corresponding frequency is given by ν = \(\frac {c}{λ}\),
 KE_max1 = h(\(\frac {c}{λ}\)) - φ 
In the second case, the incident light has a wavelength of \(\frac {λ}{2}\). The corresponding frequency is ν = \(\frac {c}{(λ/2)}\) = \(\frac {2c}{λ}\) 
Thus, the maximum kinetic energy (KE_max2) in the second case can be expressed as: 
KE_max2 = h(\(\frac {2c}{λ}\)) - φ 
Given that KE_max1 is one-third of KE_max2, we can write the equation: 
KE_max1 = \(\frac {1}{3}\)KE_max2 
Substituting the expressions for KEmax1 and KE_max2, we have: 
h(\(\frac {c}{λ}\)) - φ = \(\frac {1}{3}\)[h(\(\frac {2c}{λ}\)) - φ] 
Simplifying this equation yields: 
h(\(\frac {c}{λ}\)) - φ = \(\frac {4}{3}\)(h(\(\frac {2c}{λ}\))) - \(\frac {1}{3}\)φ h(\(\frac {c}{λ}\)) - φ = \(\frac {4}{3}\)(h(\(\frac {c}{λ}\))) - \(\frac {1}{3}\)φ 
Rearranging the equation, we find: 
h(\(\frac {c}{λ}\)) - \(\frac {4}{3}\)(h(\(\frac {c}{λ}\))) = φ - \(\frac {1}{3}\)φ 
Multiplying through by 3, we get: 
3h(\(\frac {c}{λ}\)) - 4h(\(\frac {c}{λ}\)) = 3φ - φ h(\(\frac {c}{λ}\)) = 2φ 
Rearranging further, we have: 
φ = -\(\frac {{h(\frac {c}{λ})}}{2}\)
Simplifying this expression, we find: φ = -\(\frac {hc}{2λ}\)
Therefore, the work function of the surface of the material is given by option (B) \(\frac {hc}{2λ}\).